Find the general solution $u(x,y,z)$ of:
$u_{x}+u_{y}+u_{z}=0$
Using the method of charactertistics.
I tried to set up the characteristic equations:
$x'(s)=1, y'(s)=1,z'(s)=1,w'(s)=0$
Where $w(s)=u(x(s),y(s),z(s))$ (being $x,y,z$ the characteristics). However, I'm not sure on how to proceed from here. I've only dealt with the method of characteristics in PDEs with initial conditions, but not in a more general manner. Intuitively, I would say that the general solution is of the form $f(x-z,y-z)$ for $f$ a function of two variables, but I don't know how to justify this.
Any help would be appreciated.
First, it's not hard to see your characteristic curves foliate $\mathbb{R}^3$. So, let us begin by having some generic initial conditions say $x(0) = x_0, y(0) = y_0, z(0) = 0$ and $u(x, y, 0) = f(x, y)$ for some sufficiently nice $f$. Then it follows \begin{align} x(t) = t+x_0, \ \ y(t) = t+y_0 \ \ \text{ and }\ \ z(t) = t. \end{align} Hence your pde, can now be rewritten as the an ode \begin{align} \frac{d}{dt}u(x(t), y(t), z(t)) = u_x+ u_y+u_z = 0 \end{align} which means \begin{align} u(x, y, z) = u(t+x_0, t+y_0, t)= u(x_0, y_0, 0)= f(x_0, y_0). \end{align} But $x_0 = x-t = x-z$ and $y_0 = y-t= y-z$ which means \begin{align} u(x, y, z) = f(x-z, y-z). \end{align}